This publication constitutes the refereed lawsuits of the twelfth foreign convention on man made Intelligence and Symbolic Computation, AISC 2014, held in Seville, Spain, in December 2014. The 15 complete papers provided including 2 invited papers have been conscientiously reviewed and chosen from 22 submissions.

This ebook constitutes the refereed lawsuits of the 3rd foreign convention on Statistical Language and Speech Processing, SLSP 2015, held in Budapest, Hungary, in November 2015. The 26 complete papers provided including invited talks have been conscientiously reviewed and chosen from seventy one submissions.

Remove all states unreachable from q0 (via whatever first search). Then L(M ) is finite if and only if the reduced DFA is a dag; this condition can be checked by depth-first search. Alternatively, but less usefully, L(M ) is finite if and only if L(M ) contains no string w such that n ≤ |w| < 2n. • Is L(M) = Σ∗ ? Remove all states unreachable from q0 (via whatever first search). Then L(M ) = Σ∗ if and only if every state in M is an accepting state. • Is L(M) = L(M )? Build a DFA N such that L(N ) = L(M ) \ L(M ) using a standard product construction, and then check whether L(N ) = ∅.

From the previous example, we know that there is a three-state DFA M11 that accepts the set of strings with the substring 11 and a nearly identical DFA M00 that accepts the set of strings containing the substring 00. By identifying the accept state of M00 with the start state of M11 , we obtain a five-state DFA that accepts the set of strings with 00 before 11. Finally, by inverting which states are accepting, we obtain the DFA we want. 1 0 0,1 1 0 0 0 0,1 1 1 1 0 1 0 0 0,1 0 1 1 1 0 1 0 0 0,1 0 1 1 Building a DFA for the language of strings in which every 00 is after every 11.